📂Number Theory

Multiplicative Function's Abelian group

Theorem ¹

The set of multiplicative functions $M$ and the binary operation $\ast$ satisfy that $(M,*)$ is an Abelian group.

Description

Just as the set of arithmetic functions $A$ becomes an Abelian group $(A,*)$ with convolution $\ast$, so does the set of multiplicative functions. Of course, $M \le A$, that is $M$ becomes a subgroup of $A$.

Proof

A $\left< G, \ast\ \right>$, a monoid, is defined as a group if there exists an $a '$ satisfying $a \ast\ a’ = a’ \ast\ a = e$ with respect to an element $a$ and an identity element $e$. That is, a group is a binary operation structure that satisfies the following properties:

• (i): The associative law holds for the operation.
• (ii): An identity element exists for every element.
• (iii): An inverse exists for every element.

Additionally, if the following condition is satisfied, it is called an Abelian group.

• (iv): The commutative law holds for the operation.

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Part (i), (iv). Associative and Commutative Laws

Properties of Convolution

• Associative Law: $\left( f \ast g \right) \ast k = f \ast (g \ast k)$
• Commutative Law: $f \ast\ g = g \ast\ f$

Multiplicative functions are arithmetic functions, and all arithmetic functions satisfy the associative and commutative laws.

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Part (ii). Identity Element

Identity: The arithmetic function $I$ defined as follows is called the identity function. $$ I(n) := \left[ {{ 1 } \over { n }} \right] $$

$$ I(mn) = I(m) I(n) = \begin{cases} 1 &, m = n = 1 \\ 0 & , \text{otherwise} \end{cases} $$ Therefore, it is a completely multiplicative function, and naturally, $I \in M$ is true. The identity $M$ of all arithmetic functions exists as the identity element that satisfies $( M,*)$. $$ I \ast\ f = f \ast\ I = f $$

Multiplicative Properties with respect to Convolution:

• [1]: If $f$ and $g$ are multiplicative functions, then $f \ast\ g$ is also a multiplicative function.
• [2]: If $g$ and $f \ast g$ are multiplicative functions, then $f$ is also a multiplicative function.

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Part (i). Closed under $\ast$

According to the multiplicative properties of convolution [1], $M$ is closed under $\ast$.

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Part (iv). Inverse

Properties of Multiplicative Functions: If $f$ is multiplicative, then $f(1) = 1$ holds true.

Inverse with respect to Convolution: If the arithmetic function $f$ satisfies $f(1) \ne 0$, then its inverse $f^{-1}$ uniquely exists.

Multiplicative function $f$ is an arithmetic function, and due to the properties of multiplicative functions, it satisfies $f(1) \ne 0$; thus, its inverse $f^{-1}$ exists. Meanwhile, as in Part (iii)., it was $f \ast\ f^{-1} = I \in M$, and according to the multiplicative properties of convolution [2], a $g = f^{-1}$ that satisfies $f \ast\ g = I$ must be a multiplicative function. In other words, a unique inverse $f^{-1} \in M$ exists for the multiplicative function $f \in M$.

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